Problem: $f(x) = -2x+7$ $g(t) = 2t-2(f(t))$ $h(n) = -4n^{2}-6n+g(n)$ $ f(h(0)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(0)$ . Then we'll know what to plug into the outer function. $h(0) = -4(0^{2})+(-6)(0)+g(0)$ To solve for the value of $h$ , we need to solve for the value of $g(0)$ $g(0) = (2)(0)-2(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = (-2)(0)+7$ $f(0) = 7$ That means $g(0) = (2)(0)+(-2)(7)$ $g(0) = -14$ That means $h(0) = -4(0^{2})+(-6)(0)-14$ $h(0) = -14$ Now we know that $h(0) = -14$ . Let's solve for $f(h(0))$ , which is $f(-14)$ $f(-14) = (-2)(-14)+7$ $f(-14) = 35$